CHM172 Physical Chemistry II - Molecules in Motion
Module 4: Diffusion
Marvin Jose F. Fernandez
Mindanao State University - Iligan Institute of Technology
September 20, 2020
CHM172 Physical Chemistry II - Molecules in Motion
Module Outline
Intended Learning Outcome
Diffusion
The thermodynamic view
The diffusion equation
The statistical view
References
CHM172 Physical Chemistry II - Molecules in Motion
Intended Learning Outcome
Module Overview
Intended Learning Outcome:
Students are able to describe how solute molecules and ions move in liquids.
Required Reading (Textbook):
Latest edition of P. W. Atkins, J. de Paula. Physical Chemistry. Oxford
University Press.
Activity/Assessment:
Problem solving exercises and Problem Set.
Module Objectives:
Students are able to:
1. describe the bulk motion of regions of a fluid.
2. derive a mathematical formulation for notion the that there is a natural
tendency for concentration to become uniform.
.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
Diffusion
That solutes in gases, liquids, and solids have a tendency to spread can be
discussed from three points of view:
1. thermodynamic viewpoint.
2. through a differential equation for the change in concentration in a
region by considering the flux of material through its boundaries.
3. viewed basing on a model in which diffusion is imagined as taking place
in a series of random small steps.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
Diffusion
Recall:
J(number) = − D
dN
dx
Fick’s first law [number]
(1)
where N is the number density and D is the diffusion coefficient. In a
number of cases it is more convenient to discuss the flux in terms of the
amount of molecules and the molar concentration, c. Division by Avogadro’s
constant turns equation (1) into
J(amount) = − D
dc
dx
Fick’s first law [amount]
(2)
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
The thermodynamic view
At Tconst and pconst ,
I maximum non-expansion work that can be done by a spontaneous
process is equal to the change in the Gibbs energy
I the spontaneous process is the spreading of a solute
I the work it could achieve per mole of solute molecules can be identified
with the change in the chemical potential of the solute: dwm =dµ.
The difference in chemical potential between the locations x + dx and x is
∂µ
dµ =µ(x + dx) − µ(x) =
dx
∂x T,p
so the molar work associated with migration through dx is
∂µ
dwm =
dx
∂x T,p
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
The thermodynamic view
I work done in moving a distance dx against an opposing force F (in this
context, a molar quantity) is dwm = -F dx
I comparing the two expressions for dwm it is seen that the slope of the
chemical potential with respect to position can be interpreted as an
effective force per mole of molecules.
This thermodynamic force is written as
∂µ
F =−
Thermodynamic force [definition]
∂x T,p
(3)
There is not a real force pushing the molecules down the slope of the
chemical potential: the apparent force represents the spontaneous tendency
of the molecules to disperse as a consequence of the Second Law and the
tendency towards greater entropy.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
The thermodynamic view
In a solution in which the activity of the solute is a, the chemical potential is
µ = µ + RT lna. The thermodynamic force can therefore be written in terms
of the gradient of the logarithm of the activity:
∂µ
∂lna
F =−
= −RT
(4)
∂x T,p
∂x T,p
If the solution is ideal, a may be replaced by c/cθ , where c is the molar
concentration and cθ is its standard value (1 mol dm−3 ):
!
∂ln(c/cθ )
RT ∂c
=−
(5)
F = − RT
∂x
c
∂x T,p
T,p
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
Deriving Fick’s first law of diffusion
I thermodynamic force acts in many respects like a real physical force.
I responsible for accelerating solute molecules until the viscous drag they
experience balances the apparent driving force and they settle down to
a steady ‘drift speed’ through the medium.
By considering the balance of apparent driving force and the retarding
viscous force it is possible to derive Fick’s first law of diffusion and relate the
diffusion coefficient to the properties of the medium.
Step 1. Find an expression for the flux due to molecules moving at the drift
speed
The amount (in moles) of solute molecules that can pass through is s∆tAc.
The flux J is this number divided by the area A and by the time interval ∆t:
J(amount) =sc
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
Deriving Fick’s first law of diffusion
Step 2. Find an expression for the drift speed
The apparent driving force (per mole) acting on the solute molecules is given
by equation (5). The molecules also experience a viscous drag which is
assumed to be proportional to the speed: expressed as a molar quantity this
force is written NA f s, where f is a constant, the ‘frictional constant’,
depending on the medium. When these two forces are in balance the
molecules will be moving at the drift speed. It therefore follows that, with
R/NA = k,
NA f s = −
RT
c
hence
s=−
RT dc
kT dc
=−
NA f c dx
f c dx
I negative sign arises because the molecules are moving opposite to the
direction of increasing concentration.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
Deriving Fick’s first law of diffusion
Step 3. Combine the two expressions
Substituting the expression for the drift speed into that for the flux gives:
J(amount) =sc = −
kT dc
kT dc
c=−
f c dx
f dx
I this expression has the same form as Fick’s first law
I the flux is proportional to the concentration gradient
I the diffusion constant D can be identified as kT /f , which is the
Stokes–Einstein relation:
D=
kT
f
Stokes-Einstein relation
(6)
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
Deriving Fick’s first law of diffusion
Substituting the expression for the drift speed into that for the flux gives:
J(amount) =sc = −
kT dc
kT dc
c=−
f c dx
f dx
I the constant f in the Stokes–Einstein relation can be inferred from the
hydrodynamic result known as Stokes’ law for the viscous drag
I the magnitude of the viscous force is 6πηas for a spherical particle of
radius a; it follows that f = 6πηa
I substituting this into equation (6) gives the Stokes–Einstein
equation:
D=
kT
6πηa
Stokes-Einstein equation
(7)
This equation is an explicit relation between the diffusion coefficient and the
viscosity for a species of hydrodynamic radius a and confirms that D is
inversely proportional to η.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The thermodynamic view
Deriving Fick’s first law of diffusion
I the drift speed can be related to the diffusion constant and the
thermodynamic force
I equating the two expressions for the flux, J = sc and J = −Ddc/dx,
we obtain s = −(D/c)dc/dx.
I the concentration gradient can be expressed in terms of the
thermodynamic force, that is, equation (5) rearranged in the form
dc/dx= −cF/RT to give
DF
s=
RT
I this relation provides a way to estimate the thermodynamic force from
measurements of the drift speed and the diffusion coefficient.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The diffusion equation
The diffusion equation
Diffusion results in the modification of the distribution of concentration of
the solute (or of a physical property) as inhomogeneities disappear.
Simple diffusion
∂c
∂2c
=D 2
∂t
∂x
Diffusion equation
(8)
I one of the most important equations for discussing the properties of
fluids.
I expresses the rate of change of concentration of a species in terms of
the inhomogeneity of its concentration.
I also called ‘Fick’s second law of diffusion’
I can be derived on the basis of Fick’s first law
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The diffusion equation
The diffusion equation
Physical interpretation of the diffusion equation:
I where the curvature is positive (a
dip, Figure 1), the change in
concentration with time is positive;
the dip tends to fill.
I where the curvature is negative (a
heap), the change in concentration
with time is negative; the heap
tends to spread.
I if the curvature is zero, then the
concentration is constant in time.
I the diffusion equation can be
regarded as a mathematical
formulation of the intuitive notion
that there is a natural tendency for
the wrinkles in a distribution to
disappear.
Figure 1: The diffusion equation implies
that, over time, peaks in a distribution
(regions of negative curvature) spread
and troughs (regions of positive
curvature) fill in.
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The diffusion equation
Diffusion with convection
Convection is the bulk motion of regions of a fluid. This process contrasts
with diffusion in which molecules move individually through the fluid. The
flux due to convection can be analysed in a similar way to diffusion.
Evaluating the change in concentration due to convection
I the rate of change of concentration in terms of the net flux is:
∂c JL,conv − JR,conv
=
∂t
l
I The amount passing through a face when divided by the area of the
face and the time interval is:
Jconv =
cAv∆t
= cv
A∆t
Convective flux
(9)
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The diffusion equation
Diffusion with convection
The concentrations on the left (cL ) and right (cR ) faces of the slab are
related to the concentration at its center, c0 , by
1
∂c
1
∂c
cR =c0 + l
cL = c0 − l
2
∂x
2
∂x
From equation (9)
JL,conv − JR,conv =(cL − cR )v = −
∂c
∂x
lv
Upon substitution
∂c
=−
∂t
∂c
∂x
v
Convection
(10)
When diffusion and convection occur together, the total rate of change of
concentration in a region is the sum of the two effects, which is described by
the generalized diffusion equation:
∂c
∂2c
∂c
=D 2 − v
∂t
∂x
∂x
Generalized diffusion equation
(11)
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The diffusion equation
Solutions of the diffusion equation
The diffusion equation (equation (8)) is a second-order differential equation
with respect to space and a first-order differential equation with respect to
time. To find solutions it is necessary to know two boundary conditions for
the spatial dependence and a single initial condition for the time dependence.
I consider an arrangement in which there is a layer of a solute (such as
sugar) at the bottom of a tall beaker of water (which for simplicity may
be taken to be infinitely tall), with base area A; x is the distance
measured up from the base.
I At t = 0 it is assumed that all N0 particles are concentrated on the
yz-plane at x = 0 (Initial condition)
I The two boundary conditions are derived from the requirements that
the concentration must everywhere be finite and the total amount of
particles present is n0 (with n0 = N0 /NA ) at all times. With these
conditions,
c(x, t) =
n0
−x2 /4Dt
e
A(πDt)1/2
One-dimensional diffusion
(12)
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The diffusion equation
Solutions of the diffusion equation
I consider a sugar lump suspended in an infinitely large flask of water.
I the concentration of diffused solute is spherically symmetrical, and at a
radius r is
c(r, t) =
n0
−r 2 /4Dt
e
8(πDt)3/2
Three-dimensional diffusion
(13)
The diffusion equation is
I useful for the experimental determination of diffusion coefficients
I can be used to predict the concentration of particles (or the value of
some other physical quantity, such as the temperature in a non-uniform
system) at any location
I can also be used to calculate the average displacement of the particles
in a given time
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The statistical view
The statistical view
An intuitive picture of diffusion
I particles moving in a series of small steps
I gradually migrating from their original positions
I particle jumps through a distance d after a time τ
I total distance travelled by a particle in time t is therefore td/τ
I it is most unlikely that a particle will end up at this distance from the
origin because each jump might be in a different direction
One-dimensional random walk model
I discussion is simplified by allowing the particles to travel only along a
straight line (the x-axis)
I each step a jump through distance d to the left or to the right
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The statistical view
The statistical view
Evaluating the probability distribution for a one-dimensional random walk
Imagine that a molecule has made N steps in total, NR of which are to the
right and NL to the left.
I displacement from origin is nd with n = Nr − NL
I many sequences of individual steps can arrive at a given final
displacement
I number of these sequences is equal to the number of ways of choosing
NR steps to the right and NL = N − NR steps to the left
W =
N!
N!
=
NL !NR !
(N − NR )!NR !
I at each step, the molecule can step to the left or right, so the total
number of possible sequences of steps is 2N
I probability of achieving a final displacement nd is
P (nd) =
N!
W
=
2N
(N − NR )!NR !2N
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The statistical view
The statistical view
I expression can be simplified by taking its logarithm
lnP = lnN ! − {ln(N − NR )! + lnNR ! + ln2N }
I using Stirling’s approximation in the form
lnx! ≈ ln(2π)1/2 + (x +
1
)lnx − x
2
I to obtain
1
1
lnP = − ln(2π) 2 + N +
ln
2
1 − NR /N
1 − Nr /N
1
+NR ln
− lnNR
NR /N
2
1/2 N
I introducing a new variable µ
µ=
NR
1
−
N
2
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The statistical view
The statistical view
I the expression for ln P can then be written in terms of N and µ alone
1
1
1
1
)ln( − µ) + N (µ + )ln( − µ)
2
2
2
2
1
1
1
1
−N (µ + )ln(µ + ) − lnN (µ + )
2
2
2
2
lnP = − ln(2π)1/2 2N − (N +
I probability of taking a step to the right or to the left is the same, it is
expected that after many steps the total number to the right will be
very close to half the number of steps
I that is, NR /N ≈ 21 ; it follows that µ << 1
I expanding the logarithms and taking the antilogarithms (using
N >> 1)
2
2
2
2N +1 e−2(N −1)µ
2e−2(N −1)µ
2e−2N µ
=
≈
P =
2N (2πN )1/2
(2πN )1/2
(2πN )1/2
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The statistical view
The statistical view
I recasting the expression for the probability
1/2
2
2
2τ
e−x τ /2td
One-dimensional random walk (14)
P (x, t) =
πt
I in the present calculation the particles can migrate in either direction
from the origin
I they can be found only at discrete points separated by d instead of
being anywhere on a continuous line
I it is possible to relate the diffusion coefficient D to the step length d
and the time between jumps, τ by the Einstein–Smoluchowski
equation:
d2
D=
2τ
Einstein–Smoluchowski equation
(15)
I the equation makes the connection between the microscopic details of
particle motion and the macroscopic parameters relating to diffusion
CHM172 Physical Chemistry II - Molecules in Motion
Diffusion
The statistical view
The statistical view
I if d/τ is interpreted as vmean and d is interpreted as λ, then the
Einstein–Smoluchowski equation becomes
D=
1
1
d(d/τ ) = λvmean
2
2
I essentially the same expression as obtained from the kinetic model of
gases
I the diffusion of a perfect gas is a random walk with an average step size
equal to the mean free path
CHM172 Physical Chemistry II - Molecules in Motion
References
References I
Peter Atkins, Julio de Paula, James Keeler
Atkin’s Physical Chemistry, 11th edition.
Oxford University Press, 2018.
Andreas Hofmann
Physical Chemistry Essentials.
Springer Nature, 2018.
https://courses.lumenlearning.com/boundless-chemistry/
chapter/kinetic-molecular-theory-of-matter/